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Proving the exactness of the Runge-Kutta algorithm to the second order in step-size

  1. Feb 17, 2012 #1
    1. The problem statement, all variables and given/known data

    We're supposed to prove that

    [tex]y_{n+1} = y_n + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6}[/tex]
    [tex] k_1 = hf(t_n, y_n)[/tex]
    [tex] k_2 = hf(t_n + \frac{h}{2}, y_n + \frac{k_1}{2})[/tex]
    [tex] k_3 = hf(t_n + \frac{h}{2}, y_n + \frac{k_2}{2})[/tex]
    [tex] k_4 = hf(t_n + h, y_n + k_3)[/tex]
    [tex]f(t, y) = \frac{dy}{dt}[/tex]

    is equivalent to the Taylor expansion up to the second order in step size, h:

    [tex] y_{n+1} = y_n + h \frac{dy}{dt} + \frac{1}{2} h^2 \frac{d^2 y}{dt^2}[/tex].

    2. Relevant equations

    We can expand a function as follows:

    [tex] g(x + \Delta x, y + \Delta y) = g(x,y) + \frac{\partial g}{\partial x} \Delta x + \frac{\partial g}{\partial y} \Delta y.[/tex]

    Also, we can write the second time derivative of y as follows:

    [tex]\frac{d^2 y}{dt^2} = \frac{df}{dt} = \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} = \frac{\partial f}{\partial y} f(t,y) + \frac{\partial f}{\partial t}[/tex].

    3. The attempt at a solution

    I will use these above equations to expand k2, k3, and k4. I will show all of my work for k2, and state my results for k3 and k4 since they are calculated similarly.

    [tex]k_2 = h \left[ f(t_n, y_n) + \frac{df}{dt} \bigg|_n \frac{h}{2} + \frac{df}{dy} \bigg_n h \frac{f(t_n, y_n)}{2} \right][/tex]
    [tex] = h \left[f(t_n, y_n) + \left( \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial t} \right)_n \frac{h}{2} + \frac{\partial f}{\partial y} \bigg|_n h \frac{f(t_n, y_n)}{2} \right][/tex]
    [tex] = h \left[ f(t_n, y_n) + \frac{df}{dy} f(t_n, y_n) \bigg_n h + \frac{h}{2} \frac{\partial f}{\partial t}[/tex]
  2. jcsd
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